Mathematics · Oct / Nov 2023 · Paper 21 · Question 9

YMA01

As printed in the paper

The question as text

9. y P (9, 40) C l R x O Figure 3 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 3 shows a sketch of part of the curve C with equation 2  y x 2 9 x 13 x 0 3

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Show the answer and mark scheme

9(a)[4]
=
x
−
d
3
x
2
x
3
d
27
9
y
2
Attempts to solve
x >
0
x
>
x
>
d
8
4
3
2
2
2
3
2
9(b)[8]
x
−
9
x
+
13 d
x
=
x
−
6
x
+
13
x
(
+
c
)
3
9
9
3
2
3
Area =
2
9 x
−
6
x
+
13
x
=
117
0
into d
4
9
d
21
y
y
Substitutes
x =
9
=
x
−
=
d
3
d
2
x
2
x
x
21
109
(
)
Finds where tangent meets xaxis
0
−
40
=
−
9
=
x
x
2
21
9
3
2
1
109
3
Area of R =
−
6
2
+
13
−
×
40
×
9
−
x
x
x
9
2
21
0
1600
857
=
117
−
=
21
21
There are no marks for integrating in part (a) or differentiating in (b). This is a problemsolving question
and candidates should be awarded these marks only for work in the correct part.
(a)
1
M1: Attempts d
y
−
x with one index correct. Allow for
2
→
1
or
→
2
x
x
x
x
d
A1: d
4
9
y
=
−
o.e which may be left un simplified. The LHS is not required
x
d
x
3
2
x
3
dM1: Attempts to find where d
y
3x
x >
0
and proceeds to
2
or
>
following correct squaring
x
>
k
k
d
3
3
Also allow d
but also condone d
y
y
0
2
0
2
and proceeds to
x
k
x =
leading to
x
=
k
d
d
x
If a candidate has a correct derivative, then condone 4
9
9
...0
...
x
−
x
where ... is =, > or
3
4
2
x
9
9
A1: States
x >
or
x
o.e. following the award of both M’s. Do not accept other ranges such as x < 0
4
4
(b)
2
2
M1: Attempts to integrate
3 x
−
9
x
+
13
scored for increasing a correct index by 1 which could include
13
→
13x
dM1: Uses limits of 9 (and 0) either way around and finds area under curve between 0 and 9.
It is dependent upon the previous M mark. Award when values are embedded
A1: Achieves 117 as the area under the curve between 0 and 9.
Be careful when awarding this mark as sight of 117 does not necessarily merit this award
9
3
2
3
You must see M1 and dM1 and have correct integration. Allow for
−
6
2
+
13
9 x
x
x
0
The use of a calculator to integrate is not allowed
M1: Attempts the gradient of the tangent.
into their d
y
Substitutes
x =
9
x (from part a or restarted) which must have one index correct.
d
A correct value (or follow through value) can imply this method only if there is a d
y
x . It cannot be
d
awarded if there is not a calculation for d
y
x in the question
d
dM1: Correct method of finding where tangent meets the xaxis. Dependent upon previous M
For this to be awarded you must see all elements or their equivalent. E.g.
into their d
y
Substitution of
x =
9
x (from part a or restarted) which must have one index correct
d
use of (9, 40) with their d
y
x to form the equation of the tangent at P
d
substitution of
y = into their correctly found tangent to find where tangent crosses xaxis
0
109
A1: Correct value
x =
21
ddM1: Complete method of finding the area of R. This is dependent upon all previous M's
It is usually scored for the area under the curve between 0 and 9 − the area of the triangle.
The area of the triangle may be attempted by integration. Look for a correct attempt at the method for
2x
the equation of a tangent, integrated to a form
β with limits of the x intercept of their tangent and 9
α
+
x
A1: Area R = 857
21
...............................................................................................................................................................................
Note that there are other ways in which the area could be attempted. Look carefully at their attempt please.
ALT I: (Area under curve between 0 and x intercept) +( Area between curve and line between x intercept and 9)
109
9
21
2
2
21
109
2
2
E.g.
x
−
9
x
+
13 d
x
+
x
−
9
x
+
13
−
x
−
d
x
3
3
2
2
109
0
21
2
2
M1: As the main method. Attempts to integrate
3 x
−
9
x
+
13
, scored for increasing a correct index by 1
dM1: As main method with same dependency. In this case it would be for applying the limits 0 to k and k to 9
2
2
to the integrated
3 x
−
9
x
+
13
A1: Implied by correct final answer
M1, dM1, A1: As main method
109
''
''
9
21
2
2
21
109
2
2
ddM1:
−
9
+
13 d
+
−
9
+
13
−
''
−
'' d
x
x
x
x
x
x
x
3
3
2
2
109
0
''
''
21
Complete method of finding the area of R. This is dependent upon all previous M's. See main scheme
for general principles for awarding this mark.
A1: Area R = 857
21
ALT II: (Area between curve and line between 0 and 9) ( Area of triangle under axis)
9
2
21
109
1
109
109
2
E.g.
x
−
9
x
+
13
−
x
−
d
x
−
×
×
3
2
2
2
2
21
0
2
2
M1: As the main method. Attempts to integrate
−
9
+
13
, scored for increasing a correct index by 1
3 x
x
dM1: As main method with same dependency. Look for applying the limits 0 and 9 to the integrated
2
2
−
9
+
13
3 x
x
A1: Area between curve and line, between 0 and 9, is 729
4
M1, dM1, A1: As main method
9
2
21
109
1
109
109
2
ddM1:
x
−
9
x
+
13
−
''
x
−
'' d
x
−
×
''
'' ''
×
''
3
2
2
2
2
21
0
Complete method of finding the area of R. This is dependent upon all previous M's. See main scheme
for general principles for awarding this mark.
A1: Area R = 857
21
(
)
6

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